Question

The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s) in a steam chamber for which T[infinity]= 200°C. In the heating process, a 20-mm-thick rubber wall (assumed to be untreaded) is taken from an initial temperature of 35°C to a midplane temperature of 170°C. If steam flow over the tire surfaces maintains a convection coefficient of 200 W/m^2·K. How long will it take to achieve the desired midplane temperature?

Answer 1

Answer:

$\mathbf{t_f = 1436.96 \ sec }$

Explanation:

Given that :

The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s)

i.e

k = 0.14 W/mK

∝ = 6.35 × 10⁻⁸ m²/s

L = 0.01 m

$B_1 = \dfrac{hL}{k} \\ \\ B_1 = \dfrac{200*0.01}{0.14} \\ \\ B_1 = 14.2857$

We cannot use the model of Lumped Capacitance; SO Let assume that Fourier Number $F_o > 0.2$

⇒  $\dfrac{T_o - T_ \infty }{T_i - T_ \infty} = C_1 exp (- \zeta_i^2 *F_o)$

From Table 5.1 ; at $B_1$ = 14.2857

$C_1 = 1.265 \\ \\ \zeta_1 = 1.458 \ rad$

$\dfrac{170-200}{35-200} = 1.265 exp [ - (1.458)^2* \dfrac{ \alpha t_f}{L^2}]$

$In ( \dfrac{0.1818}{1.265}) = \dfrac{-1.458^2*6.35*10^{-8}*t_f}{0.01^2}$

$-1.9399=-0.001350 *t_f$

$t_f = \dfrac{-1.9399}{-0.001350}$

$\mathbf{t_f = 1436.96 \ sec }$